Problem: In the diagram, $P$ is on $RS$ so that $QP$ bisects $\angle SQR$.  Also, $PQ=PR$, $\angle RSQ=2y^\circ$, and $\angle RPQ=3y^\circ$.  What is the measure, in degrees, of $\angle RPQ$? [asy]
// C14
import olympiad;
size(7cm);

real x = 50; real y = 20;

pair q = (1, 0);
pair r = (0, 0);
pair p = intersectionpoints((10 * dir(x))--r, q--(shift(q) * 10 * dir(180 - x)))[0];
pair s = intersectionpoints(r--(r + 10 * (p - r)), 10 * dir(180 - 2 * x)--q)[0];

// Draw lines
draw(p--s--q--p--r--q);

// Label points
label("$R$", r, SW);
label("$Q$", q, SE);
label("$S$", s, N);
label("$P$", p, NW);

// Label angles
label("$x^\circ$", q, 2 * W + 2 * NW);
label("$x^\circ$", q, 4 * N + 2 * NW);
label("$2y^\circ$", s, 5 * S + 4 * SW);
label("$3y^\circ$", p, 4 * S);

// Tick marks
add(pathticks(r--p, 2, spacing=0.6, s=2));
add(pathticks(p--q, 2, spacing=0.6, s=2));
[/asy]
Solution: Since $RPS$ is a straight line, then $\angle SPQ = 180^\circ - \angle RPQ = 180^\circ - 3y^\circ$.

Using the angles in $\triangle PQS$, we have $\angle PQS + \angle QSP + \angle SPQ = 180^\circ$. Thus, $x^\circ+2y^\circ + (180^\circ - 3y^\circ) = 180^\circ$ or $x-y+180 = 180$ or $x=y$.

(We could have instead looked at $\angle RPQ$ as being an external angle to $\triangle SPQ$.)

Since $x=y$, then $\angle RQS=2y^\circ$.

Since $RP=PQ$, then $\angle PRQ=\angle PQR=x^\circ = y^\circ$. [asy]
// C16S
import olympiad;
size(7cm);

real x = 36; real y = 36;

pair q = (1, 0);
pair r = (0, 0);
pair p = intersectionpoints((10 * dir(x))--r, q--(shift(q) * 10 * dir(180 - x)))[0];
pair s = intersectionpoints(r--(r + 10 * (p - r)), 10 * dir(180 - 2 * x)--q)[0];

// Draw lines
draw(p--s--q--p--r--q);

// Label points
label("$R$", r, SW);
label("$Q$", q, SE);
label("$S$", s, N);
label("$P$", p, NW);

// Label angles
label("$y^\circ$", q, 4 * W + 2 * NW);
label("$y^\circ$", q, N + 5 * NW);
label("$y^\circ$", r, 2 * NE + 3 * E);
label("$2y^\circ$", s, 3 * S + SW);
label("$3y^\circ$", p, 3 * S);

// Tick marks
add(pathticks(r--p, 2, spacing=0.6, s=2));
add(pathticks(p--q, 2, spacing=0.6, s=2));

[/asy] Therefore, the angles of $\triangle RQS$ are $y^\circ$, $2y^\circ$ and $2y^\circ$.

Thus, $y^\circ+2y^\circ+2y^\circ=180^\circ$ or $5y=180$ or $y=36$.

Therefore, $\angle RPQ=3y^\circ = 3(36)^\circ=108^\circ$, so our final answer is $\boxed{108}$ degrees.